A neutron initially at rest, decays into a proton, an electron, and...

By the principle of conservation of linear momentum, the initial momentum of the neutron is equal to the vector sum of the final momenta of the decay products.

$\vec{p}_p + \vec{p}_e + \vec{p}_{\bar{\nu}} = 0$

Since the electron and the antineutrino are ejected along the same direction, their momenta add up algebraically. The magnitude of the proton's momentum is equal to the sum of their momenta:

$p_p = p_e + p_{\bar{\nu}}$

$p_p = 1.4 \times 10^{-26} + 6.4 \times 10^{-27}$

$p_p = 14 \times 10^{-27} + 6.4 \times 10^{-27} = 20.4 \times 10^{-27} \text{ kg-m/s}$

The recoil speed of the proton is given by $v_p = \dfrac{p_p}{m_p}$:

$v_p = \dfrac{20.4 \times 10^{-27}}{1.67 \times 10^{-27}}$

$v_p = \dfrac{20.4}{1.67} \approx 12.2 \text{ m/s}$